Question

Let \(A\) be a square matrix of order \(3\). If \(\det A=2\), then the value of \(\det(\operatorname{adj} A^3)\) is

• A. \(2^3\)
• B. \(2^6\)
• C. \(2^9\)
• D. \(2^{12}\)

Hint:

For an \(n\times n\) matrix \(B\), \(\det(\operatorname{adj}B)=(\det B)^{n-1}\).

Answer 1

The Correct Option is B

Concept:
For a square matrix \(B\) of order \(n\), \[ \det(\operatorname{adj}B)=(\det B)^{n-1} \] Here the order is \[ n=3 \]

Step 1: Find determinant of \(A^3\). Given, \[ \det A=2 \] Then, \[ \det(A^3)=(\det A)^3 \] \[ =2^3 \]

Step 2: Use adjoint determinant formula. Let \[ B=A^3 \] Since \(B\) is also of order \(3\), \[ \det(\operatorname{adj}B)=(\det B)^{3-1} \] \[ =(\det B)^2 \] \[ =(2^3)^2 \] \[ =2^6 \]

Step 3: Final answer. \[ \boxed{2^6} \]